If `x+ (1)/(x)= sqrt3` find
`x^(67) + x^(53) + x^(43) + x^(29) +x^(24) + x^(18) + x^(6) + 3`

Answer

To solve the problem, we start with the equation given:

\[ x + \frac{1}{x} = \sqrt{3} \]

We need to find the value of:

\[ x^{67} + x^{53} + x^{43} + x^{29} + x^{24} + x^{18} + x^{6} + 3 \]

### Step 1: Find \(x^3 + \frac{1}{x^3}\)

First, we will cube both sides of the equation \(x + \frac{1}{x} = \sqrt{3}\):

\[ \left(x + \frac{1}{x}\right)^3 = \left(\sqrt{3}\right)^3 \]

Using the identity \(a^3 + b^3 + 3ab(a + b)\):

\[ x^3 + \frac{1}{x^3} + 3\left(x \cdot \frac{1}{x}\right)\left(x + \frac{1}{x}\right) = 3\sqrt{3} \]

Since \(x \cdot \frac{1}{x} = 1\), we simplify:

\[ x^3 + \frac{1}{x^3} + 3\sqrt{3} = 3\sqrt{3} \]

This leads to:

\[ x^3 + \frac{1}{x^3} = 3\sqrt{3} - 3\sqrt{3} = 0 \]

### Step 2: Find \(x^6 + \frac{1}{x^6}\)

Next, we can use the result from \(x^3 + \frac{1}{x^3}\) to find \(x^6 + \frac{1}{x^6}\):

\[ \left(x^3 + \frac{1}{x^3}\right)^2 = x^6 + \frac{1}{x^6} + 2 \]

Substituting \(x^3 + \frac{1}{x^3} = 0\):

\[ 0^2 = x^6 + \frac{1}{x^6} + 2 \]

Thus,

\[ x^6 + \frac{1}{x^6} = -2 \]

### Step 3: Find \(x^6\)

From \(x^6 + 1 = 0\), we have:

\[ x^6 = -1 \]

### Step 4: Simplify the original expression

Now we can express each term in the original expression in terms of \(x^6\):

\[ x^{67} = x^{66} \cdot x = (x^6)^{11} \cdot x = (-1)^{11} \cdot x = -x \]
\[ x^{53} = x^{52} \cdot x = (x^6)^8 \cdot x^4 = (-1)^8 \cdot x^4 = x^4 \]
\[ x^{43} = x^{42} \cdot x = (x^6)^7 \cdot x^5 = (-1)^7 \cdot x^5 = -x^5 \]
\[ x^{29} = x^{24} \cdot x^5 = (x^6)^4 \cdot x^5 = (-1)^4 \cdot x^5 = x^5 \]
\[ x^{24} = (x^6)^4 = (-1)^4 = 1 \]
\[ x^{18} = (x^6)^3 = (-1)^3 = -1 \]
\[ x^{6} = -1 \]

Now substituting these values into the expression:

\[ -x + x^4 - x^5 + x^5 + 1 - 1 - 1 + 3 \]

### Step 5: Combine like terms

This simplifies to:

\[ -x + x^4 + 3 - 1 - 1 = -x + x^4 + 1 \]

### Step 6: Substitute \(x + \frac{1}{x} = \sqrt{3}\)

To find \(x^4\), we can use the relation \(x^2 + \frac{1}{x^2}\):

\[ \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} = 3 \]
\[ x^2 + \frac{1}{x^2} = 3 - 2 = 1 \]

Now, we can find \(x^4 + \frac{1}{x^4}\):

\[ \left(x^2 + \frac{1}{x^2}\right)^2 = x^4 + 2 + \frac{1}{x^4} = 1 \]
\[ x^4 + \frac{1}{x^4} = 1 - 2 = -1 \]

Thus, \(x^4 = -1\).

### Final Calculation

Substituting back into our expression:

\[ -x + (-1) + 1 = -x + 0 = -x \]

Since \(x + \frac{1}{x} = \sqrt{3}\), we can conclude:

\[ -x = -\left(\frac{\sqrt{3} - 1}{2}\right) \]

Thus, the final answer is:

\[ \boxed{2 - 2\sqrt{3}} \]